ĐK: \(3x\ne\pm y;x\ne0\)

A = \(\dfrac{3x}{3x+y}-\dfrac{x}{3x-y}+\dfrac{2x}{\left(3x-y\right)\left(3x+y\right)}\)

= \(\dfrac{3x\left(3x-y\right)-x\left(3x+y\right)+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{6x^2-4xy+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{2x\left(3x-2y+1\right)}{\left(3x-y\right)\left(3x+y\right)}\)

Thay x = 1; y=2, ta có:

A = \(\dfrac{2.1\left(3.1-2.2+1\right)}{\left(3.1-2\right)\left(3.1+2\right)}=0\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow x=2k;y=3k\)

\(P=\dfrac{4k^2-2k.3k+9k^2}{4k^2+2k.3k+9k^2}=\dfrac{13k^2-6k^2}{13k^2+6k^2}=\dfrac{7k^2}{19k^2}=\dfrac{7}{19}\)

\(A=\dfrac{2x^2\left(3x-4y+2\right)}{x\left(3x+y\right)\left(3x-y\right)}=\dfrac{2x\left(3x-4y+2\right)}{\left(3x+y\right)\left(3x-y\right)}\\ A=\dfrac{2\left(3-8+2\right)}{\left(3+2\right)\left(3-2\right)}=\dfrac{2\left(-3\right)}{5}=\dfrac{-6}{5}\)

\(a,\dfrac{3x+21}{x^2-9}+\dfrac{2}{x+3}-\dfrac{3}{x-3}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}-\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21+2x-6-3x-9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2}{x-3}\)

\(b,\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\\ =\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x+3}{x^2-1}\\ =\dfrac{3x^2+4x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2-2x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{3x^2+4x+1-x^2+2x-1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2+2x-3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{2x^2+6x-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x^2+3x\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)

\(=\dfrac{x\left(x+3\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x+3}{\left(x-1\right)^2}\)

\(\dfrac{x^2-2xy+y^2}{x^2-xy}=\dfrac{\left(x-y\right)^2}{x\left(x-y\right)}=\dfrac{x-y}{x}\)

\(\dfrac{x^2-4}{3x+6}=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+2\right)}=\dfrac{x-2}{3}\)

a. \(\dfrac{x^2-2xy+y^2}{x^2-xy}=\dfrac{\left(x-y\right)^2}{x\left(x-y\right)}=\dfrac{x-y}{x}\)

b. \(\dfrac{x^2-4}{3x+6}=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+2\right)}=\dfrac{x-2}{3}\)

  \(\dfrac{3x+2}{x^2-2x+1}-\dfrac{6}{x^2-1}-\dfrac{3x-2}{x^2+2x+1}\)

= \(\dfrac{3x+2}{\left(x-1\right)^2}-\dfrac{6}{\left(x-1\right)\left(x+1\right)}-\dfrac{3x-2}{\left(x+1\right)^2}\)

= \(\dfrac{\left(3x+2\right)\left(x+1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}-\dfrac{6\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)^2}-\dfrac{\left(3x-2\right)\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}\)

= \(\dfrac{3x^3+8x^2+7x+2}{\left(x^2-1\right)^2}-\dfrac{6x^2-6}{\left(x^2-1\right)^2}-\dfrac{3x^3-8x^2+7x-2}{\left(x^2-1\right)^2}\)

= \(\dfrac{10x^2+10}{\left(x^2-1\right)^2}\)

= \(\dfrac{10\left(x^2+1\right)}{\left(x^2-1\right)^2}\)

\(E=\left[\left(\dfrac{3}{x+1}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)

\(=\left[\left(\dfrac{3}{x+1}-\dfrac{x}{\left(x+1\right)^2}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)

\(=\left[\left(\dfrac{2x+3}{\left(x+1\right)^2}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)

\(=\left[\dfrac{x^2+7x}{x\left(x+1\right)^2}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)

\(=\dfrac{2x\left(2x+5\right)}{x\left(x+1\right)^2}.\dfrac{x^2+x}{3x+1}\)

\(=\dfrac{2x\left(2x+5\right)}{x\left(x+1\right)^2}.\dfrac{x^2+x}{3x+1}=\dfrac{2x\left(2x+5\right)}{\left(x+1\right)\left(3x+1\right)}\)

\(A=\left(\dfrac{\left(x-1\right)^2}{x^2+x+1}+\dfrac{2x^2-4x-1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right)\cdot\dfrac{x^2+1}{x+1}\)

\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{x+1}\)

\(=\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{x+1}=\dfrac{x^2+1}{x+1}\)

a, Với x khác 1 

\(A=\dfrac{x^2+x+1-3x^2+2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}=-\dfrac{1}{x^2+x+1}\)

b, Ta có \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Rightarrow\dfrac{-1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< 0\)

Vậy với x khác 1 thì bth A luôn nhận gtri âm